【伪语法基础】for循环练习2
令00=1,fx=i=0∑nix×yi 。
则有:
\begin{align} f_x+(n+1)^x\times y^{n+1}\;&=\;\sum_{i=0}\limits^n(i+1)^n\times y^{i+1}+0\\ &=\sum_{i=0}\limits^n \bigg[\;y^{i+1}\big( \sum_{j=0}\limits^xC^j_x\times i^j \big) \bigg]\\ &=\sum_{i=0}\limits^n\sum_{j=0}\limits^xy^{i+1}i^jC^j_x\\ &=y\times\; \sum_{j=0}\limits^x \bigg(\; \sum_{i=0}\limits^ni^j\times y^i \; \bigg)\\ &=y\times\; \sum_{j=0}\limits^x C^j_xf_j \end{align}
∴fj=((n+1)x×yn+1−y×j=0∑xCxjfj)×(y−1)−1[y=1]
此时,若y==1,则有:
\begin{align} (n+x)^x\times y^{n+1}\;&=\;y\times \sum_{j=0}\limits^{x-2} C^j_xf_j\,+\,y\times C^{x-1}_x\times f_{x-1}\\ &令\;x=x+1\;,得\text:\\ (n+x)^{x+1}\times y^{n+1}\;&=\;y\times \sum_{j=0}\limits^{x-1} C^j_xf_j\,+\,y\times C^{x}_{x+1}\times f_x\\ &进而得\text :\\ f_x\;&=\;\bigg[\;\; (n+x)^{x+1}\times y^{n+1}\,-\,\sum_{j=0}\limits^{x-1} y\,C^j_xf_j \;\;\bigg]\,\times [\, y\times C^x_{x+1} \,]^{-1} \end{align}